Java - Transfer the data from one list into another list

Question

Here is the scenario, in my source list, it contains all the Users object. Each user object will have id, event, and timestamp. I need to create a destination list to contains all the user object that which have latest timestamp record for each id. Like example below

import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Date;
import java.util.List;
import com.vincent.object.User;

public class Test {
  public static void main(String[] args) throws Exception {
    DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.SSS");
    User u1 = new User("1", "55", dateFormat.parse("2017-10-01 10:11:01.111"));
    User u2 = new User("1", "105", dateFormat.parse("2017-10-01 10:11:02.111"));
    User u3 = new User("2", "55", dateFormat.parse("2017-10-01 10:11:03.111"));
    List<User> sources = new ArrayList<>();
    sources.add(u1);
    sources.add(u2);
    sources.add(u3);

    List<User> destination = new ArrayList<>();
    // I want my destination array only contains following 2 result:
    // u2 and u3 from the source
  }
}

How can I approach this?

EDIT: Here is the User class

import java.util.Date;

public class User {
  private String id;
  private String reason;
  private Date date;

  public User(String id, String reason, Date date) {
    super();
    this.id = id;
    this.reason = reason;
    this.date = date;
  }
  // getter setter
}

If I read the question, the latest date of user "1", and user "2". Some stream operation? — M. le Rutte, Oct 12 '17 at 15:34
Hi Vincent do you mean hold one list in another ?Can't you just store source list in destination after adding and add a condition to check whether contents of the list a re same by using !Collections.disjoint(list1, list2); or normal equals method list1.equals(list2); — Pradeep, Oct 12 '17 at 15:36
Here is a question with stream that uses a stateful filter predicate: https://stackoverflow.com/questions/23699371/java-8-distinct-by-property — M. le Rutte, Oct 12 '17 at 15:39
FYI, `Date` and `SimpleDateFormat` are part of the troublesome old date-time classes that are now legacy, supplanted by the java.time classes. — Basil Bourque, Oct 12 '17 at 16:21

score 3 · Accepted Answer · answered Oct 12 '17 at 15:58

You can use Java 8 streams for this.

public class Test {
    public static void main(String[] args) throws Exception {
        DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.SSS");
        User u1 = new User("1", "55", dateFormat.parse("2017-10-01 10:11:01.111"));
        User u2 = new User("1", "105", dateFormat.parse("2017-10-01 10:11:02.111"));
        User u3 = new User("2", "55", dateFormat.parse("2017-10-01 10:11:03.111"));
        User u4 = new User("2", "105", dateFormat.parse("2017-10-01 10:11:04.111"));
        List<User> sources = new ArrayList<>();
        sources.add(u1);
        sources.add(u2);
        sources.add(u3);
        sources.add(u4);


        List<User> destination = sources.stream()
                .collect(Collectors.groupingBy(User::getId, Collectors.maxBy(Comparator.comparing(User::getDate))))
                .values()
                    .stream()
                    .map(o -> o.get())
                    .collect(Collectors.toList());

        System.out.println(destination);
    }
}

Here I'm gruping by Ids and getting the user with the max date.

Output (obviously overriding toString on User class):

[User{id='1', reason='105', date=Sun Oct 01 10:11:02 BOT 2017},
User{id='2', reason='105', date=Sun Oct 01 10:11:04 BOT 2017}]

score 0 · Answer 2 · answered Oct 12 '17 at 15:48

I'm assuming you don't need to keep the order of the users as it was in the source list. The idea is the following:

iterate over the source list
add each user in a map if it was not present, update the user if the one stored had a timestamp lower than the current one (assuming that two users are equals if they have the same id)

return the key set of the map.

public static void main(String[] args)
{
     DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.SSS");
     User u1 = new User("1", "55", dateFormat.parse("2017-10-01 10:11:01.111"));
     User u2 = new User("1", "105", dateFormat.parse("2017-10-01 10:11:02.111"));
     User u3 = new User("2", "55", dateFormat.parse("2017-10-01 10:11:03.111"));
     List<User> sources = new ArrayList<>();
     sources.add(u1);
     sources.add(u2);
     sources.add(u3);

     Set<User> uniqUsers = new HashSet<>();
     for(User u : sources)
     {
         if(uniqUsers.containsKey(u))
         {
             User oldUser = uniqUsers.get(u);
             long oldTimeStamp = uniqUsers.get(u).getDate().getTime();
             long currentTimeStamp = u.getDate().getTime();
             if(currentTimeStamp > oldTimeStamp)
                 uniqUsers.remove(oldUser);
            uniqUsers.add(u);
         }
      }
      List<User> destination = new ArrayList<>(uniqUsers);

 }

The User class must be modified in this way (in order to work correctly with the HashSet.

public class User 
{
    private String id;
    private String reason;
    private Date date;

    public User(String id, String reason, Date date) 
    {
        super();
        this.id = id;
        this.reason = reason;
        this.date = date;
     }
     // getter setter


     @Override
     public int hashCode()
     {
        return id.hashCode();
     }

     @Override
     public boolean equals(Object obj)
     {
         return obj instanceof User && ((User) obj).id.equals(id);
     }
}

if you want to use streams as suggested in some comments you can always applying a filter that implement the HashSet logic. Another solution could be to sort the array by date (`Collections.sort(sources, Comparator.comparingInt(u -> u.getDate().getTime())`); and then iterating from right to left keep each user only the first time you meet him. — Tommaso Pasini, Oct 12 '17 at 15:52

Java - Transfer the data from one list into another list

2 Answers2